Partition function

[apricimo]

If I have a partition function say p = 1 + Kx where K is some constant describing a simple ligand binding to a macromolecule and x is the concetration of the ligand. What is the significance of taking the derivative of this function such that you take the natural log of the function and take the derivative with resepect to the natural log of x, so dln(p)/dln(x). This give me X which is the average ligation of the system (i.e. 0.5 or 0.2 of the macromolecules have a ligand bound). Where do these natural logs come from?

 

Anyone have an idea what thats all about?

[timo]

EDIT: After writing the following paragraphs I noticed that the expressions you wrote do not fit what I said. Maybe your expressions are wrong or there's another obscuring step happening - but maybe there is something else that is called partition function that is different from the partition function in statistical mechanics.

 

 

I only know partition functions from statistical mechanics. There, the reason why the logs and the derivatives appear is that the partition function is a sum over exponential functions which have products of the observables and their conjugates (under a Legendre transformation) in their exponent. Since that probably does not help you much let's give an example:

 

 

Take a system at temperature [math] T = \frac{1}{k\beta} [/math] (the point of the equation is to define beta which I will use later) which can have energies E. The number of possibilities for a system to have energy E is the density of states g(E). The partition function Z is [math] Z = \int_{-\infty}^{\infty} \, dE \ g(E) e^{-\beta E} [/math]. The probability that the system has an energy between [math]E_1[/math] and [math]E_2[/math] is [math]\frac 1Z \int_{E_1}^{E_2} \, dE \ g(E) e^{-\beta E}[/math] so in that sense the partition function is just a normalization constant for computing probabilities. The expectancy value [math] \left< X \right> [/math] for some observable X then is [math] \left< X \right> = \frac 1Z \int_{-\infty}^{\infty} \, dE \ X(E) g(E) e^{-\beta E} [/math], i.e. the value of the observable at all energies weighted by the probability to encounter that energy (this requires the observable to be a function of energy, of course).

Now, here comes the trick: If I want to compute the average energy of the system then the calculation can be simplified:

From definition [math] \left< E \right> = \frac 1Z \int_{-\infty}^{\infty} \, dE \ g(E) E e^{-\beta E}[/math]

Since taking a derivative of the exponential function with respect to beta would bring a factor of E in front of it this can be rewritten as

[math] = \frac 1Z \int_{-\infty}^{\infty} \, dE \ g(E) \frac{-\partial e^{-\beta E}}{\partial \beta} [/math]

Pulling the derivative in front of the integral

[math]= \frac {-1}{Z} \frac{\partial}{\partial \beta} \int_{-\infty}^{\infty} \, dE \ g(E) e^{-\beta E} [/math]

The integral of course just euqals Z, so

[math]= \frac {-1}{Z} \frac{\partial Z}{\partial \beta}[/math]

And from using the chain-rule backwards this can be rewritten as

[math] = - \frac{\partial \ln Z}{\partial \beta} [/math]

And since you are probably more used to temperatures this rewrites as

[math] = \frac{1}{kT^2} \frac{\partial \ln Z}{\partial T}[/math]

 

So to answer your questions from that:

- The derivative appears when eliminating the variable to average over under the integral.

- The logarithm appears when absorbing the normalization constant into the derivative.

 

 

Of course, taking E as the variable and [math]\beta[/math] as its conjugate was just an example. You can have any other pair of variables related in this way and the last step (rewriting into variables that do not appear as a simple product XY in the exponent) might obscure the result in favor of more experiment-related variables. But the basic scheme is always the one above.

[apricimo]

Cool I like your explanation.. but taking a derivative with respect to ln(x) that part I don't understand. I understand taking a ln of a function to make a derivative simpler but what if you had y = 1 + Kx and the derivative you take is dln(y)/dln(x)...